A drone

A flying drone aimed the area for an architect. He took off perpendicularly from point C to point D. He was at a height of 300 m above the plane of ABC. The drone from point D pointed at a BDC angle of 43°. Calculate the distance between points C and B in meters.

Result

x =  279.755 m

Solution:

$CD = 300 \ m \ \\ \angle BDC = 43 \ ^\circ \ \\ \angle DCB = 90 \ ^\circ \ \\ \ \\ δ = \angle BDC ^\circ \rightarrow rad = \angle BDC ^\circ \cdot \ \dfrac{ \pi }{ 180 } \ = 43 ^\circ \cdot \ \dfrac{ 3.1415926 }{ 180 } \ = 0.75049 \ \\ \ \\ \tan δ = BC / CD \ \\ \tan δ = x / CD \ \\ \ \\ x = CD \cdot \ \tan ( δ ) = 300 \cdot \ \tan ( 0.7505 ) \doteq 279.7545 = 279.755 \ \text{ m }$

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